Why a Narrowing Pipe Changes Fluid Speed
A fluid passing through a narrower part of a pipe has less cross-sectional area available to pass through. If the same volume of fluid must pass each section in the same amount of time, the fluid must move faster through the narrower section.
This relationship is an application of conservation of volume flow. It connects a geometric quantity, cross-sectional area, to a motion quantity, flow speed. When area changes, differentiating the same relationship also gives the rate at which speed changes.
Volume Flow Rate and the Continuity Relation
The volume flow rate \(Q\) is the volume of fluid passing through a cross-section per unit time. Its units are volume divided by time, such as \(\text{m}^3/\text{s}\) or \(\text{cm}^3/\text{s}\). For fluid moving uniformly across a cross-section of area \(A\) at speed \(v\) perpendicular to that section, the flow rate is \(Q=Av\).
For a nonuniform velocity across the section, \(v\) in this formula is understood as the average velocity normal to the cross-section. In either case, the formula describes the volume crossing that section per unit time.
The assumptions matter. Steady flow means the flow at a given location does not change with time. Incompressibility means the fluid’s density is constant, so conserving mass also conserves volume. If fluid can accumulate, escape, or enter through a branch, the flow rates at the two sections need not be equal.
Why Area Times Speed Gives the Flow Rate
Imagine fluid passing through a cross-section of area \(A\) at a uniform speed \(v\). Over a short time interval \(\Delta t\), the fluid travels a distance \(v\Delta t\). The volume that crosses the section is the area multiplied by this distance:
Dividing by \(\Delta t\) gives the volume per unit time, \(Q=\Delta V/\Delta t=Av\). If the flow is steady and incompressible, the same volume that crosses one section must cross every other section during the same time interval. Thus \(A_1v_1=A_2v_2\). This explains why a decrease in area requires an increase in speed when the flow rate is fixed.
Comparing Speeds at Two Pipe Sections
The continuity relation can be rearranged to find an unknown speed. For example, \(v_2=(A_1/A_2)v_1\). Since the ratio of the areas is dimensionless, \(v_2\) has the same units as \(v_1\). For a circular pipe, \(A=\pi r^2\), so the area depends on the square of the radius. A pipe with half the radius has one-quarter the area, not half the area.
Worked Example: Speed in a Pipe with Half the Radius
A pipe narrows from radius \(4\text{ cm}\) to radius \(2\text{ cm}\). The fluid’s speed in the wider section is \(1.5\text{ m/s}\). Find its speed in the narrower section, assuming steady incompressible flow with no leaks.
First find the ratio of the cross-sectional areas. The factors of \(\pi\) cancel:
Apply \(A_1v_1=A_2v_2\) and solve for \(v_2\):
Check the flow rates in consistent units. The radii are \(0.04\text{ m}\) and \(0.02\text{ m}\), so
The flow rates agree, as required. The narrower section’s speed is \(6\text{ m/s}\).
Finding a Speed Rate as Area Changes
Suppose the cross-sectional area \(A\) and speed \(v\) vary with time, but the volume flow rate \(Q\) is held constant. Then \(A(t)v(t)=Q\). Differentiating both sides with respect to time gives the related-rates equation
Both factors in the product can change, so both terms must appear. Solving for the speed rate gives
When the area is decreasing, \(dA/dt<0\), while \(A\) and \(v\) are positive. The formula then gives \(dv/dt>0\): the speed is increasing. If the area is increasing while flow rate stays fixed, the speed decreases.
Worked Example: Speeding Up as Area Decreases
At an instant, a pipe section has area \(12\text{ cm}^2\) and fluid speed \(5\text{ cm/s}\). The area is decreasing at \(3\text{ cm}^2/\text{s}\). If the volume flow rate is constant, find the speed rate at that instant.
Because area is decreasing, its rate is \(dA/dt=-3\text{ cm}^2/\text{s}\). Substitute the values into the rate formula:
The positive answer means the speed is increasing. Check by differentiating the product at this instant:
The derivative of the flow rate is zero, as required for constant flow. The calculated speed rate is \(1.25\text{ cm/s}^2\).
When the Pipe Narrows Along Its Length
A pipe may have a fixed, narrowing shape rather than an area that changes over time. In that situation, describe the area and speed as functions of position \(x\) along the pipe. For steady incompressible flow, \(A(x)v(x)=Q\) at every position. Differentiating with respect to \(x\), rather than time, gives
This is a spatial rate: it describes how speed changes per unit distance along the pipe. It is not the same quantity as \(dv/dt\). If the pipe is circular and its radius \(r\) changes with \(x\), use \(A=\pi r^2\) and differentiate to obtain \(dA/dx=2\pi r\,dr/dx\). Consequently, \(dv/dx=-(2v/r)(dr/dx)\).
Worked Example: Speed Change Along a Tapered Pipe
At a point in a circular pipe, the radius is \(0.05\text{ m}\), the speed is \(2\text{ m/s}\), and the radius decreases in the direction of flow at a rate \(dr/dx=-0.01\). Find \(dv/dx\) at that point. Assume steady incompressible flow with no leaks.
Since \(A=\pi r^2\), differentiate with respect to \(x\):
The area at the point is \(A=\pi(0.05\text{ m})^2=0.0025\pi\text{ m}^2\). Use the spatial rate relation:
The positive sign indicates that speed increases as \(x\) increases in the direction of flow. Check the differentiated flow relation:
Thus the local derivative is \(0.8\text{ s}^{-1}\), or \(0.8\text{ m/s}\) of speed increase per metre along the pipe.
A Common Sign and Product-Rule Mistake
It is easy to reverse the sign in the time-rate formula and conclude that a narrowing pipe slows the fluid. For the values in the previous time-rate example, a mistaken calculation might be
This is incorrect: it predicts that speed decreases as area decreases, despite the flow rate being fixed. Substituting this answer into the product rule exposes the error:
The correct sign follows directly from differentiating \(Av=Q\): \(A\,dv/dt+v\,dA/dt=0\). Moving the second term to the other side gives \(A\,dv/dt=-v\,dA/dt\), so \(dv/dt=-(v/A)(dA/dt)\). The negative sign is essential.
A second frequent error is to use area in direct proportion to radius. For a circle, \(A=\pi r^2\); halving the radius quarters the area. Before comparing speeds, calculate or simplify the ratio of areas, not just the ratio of radii.
How to Set Up a Pipe-Flow Rate Problem
Check that the problem gives steady incompressible flow with no leaks or branches, or explicitly states that \(Q\) is constant.
Use \(A_1v_1=A_2v_2\) for two sections, or \(A(t)v(t)=Q\) for a time-dependent area at fixed flow rate.
For a circular pipe, use \(A=\pi r^2\); do not replace area with radius.
Apply the product rule when both area and speed can change. Use time for a temporal rate or position for a spatial rate.
Verify the differentiated conservation equation, interpret the sign, and confirm that the units match the rate being asked for.
Why the Relationship Is Useful
The area-speed relationship turns a geometric change into a prediction about motion. It explains, under the stated assumptions, why fluid speed is higher in a narrower section and provides a quantitative way to compare speeds or calculate a local speed rate. It also shows why signs and units matter: a negative area rate and a positive speed rate are consistent with constant flow, while a result with the wrong sign fails the conservation equation.
The relation is conditional, not a claim that every real pipe has the same flow rate at every instant. If the fluid is compressible, if flow changes with time, or if fluid is added or removed between sections, additional information is needed. Use \(Av=Q\) as a conservation statement only when its assumptions fit the situation.
Check Your Understanding
Use the continuity relation and its assumptions to answer these questions.
- A pipe’s area at one section is three times the area at another. If the flow rate is constant, how do the speeds at the two sections compare?
- For a circular pipe, what happens to the cross-sectional area when the radius is reduced to one-third of its original value?
- If \(A=10\text{ cm}^2\), \(v=4\text{ cm/s}\), and \(dA/dt=-2\text{ cm}^2/\text{s}\) at constant flow rate, find \(dv/dt\). Include units.
- Why must differentiating \(A(t)v(t)=Q\) include two terms?
- What does a positive value of \(dv/dx\) mean for flow along the direction in which \(x\) increases?
